Laplace Transform of y''(t)-y(t)=0 w/ Initial Conditions

[Ry122]

I need to find the laplace transform of
y''(t)-y(t)=0 with initial conditions y(0)=1 and y'(0)=3

My attempt:
Ly''(t)=(s^2)F(s)-s(1)-(3)
Ly'(t)=sF(s)-1
so (s^2)F(s)-s(1)-(3)-sF(s)-1=0
I need to isolate F(s) so
F(s)=-(2/s^2)
Is this correct?

 

[Cyosis]

You're solving for a different differential equation than the one you have listed. You are solving $y''(t)-y'(t)=0$. That said you still made some mistakes. Your first two steps are correct, but then when you subtract the two you make a minus sign error. It should be,

$$s^2 F(s)-s-3-(s F(s)-1)=s^2 F(s)-s-3-sF(s) \textcolor{red}{+} 1=0$$.

I am not sure how you isolated the F(s), something seems to have gone wrong there as well. You can write it like,

$$(s^2-s)F(s)-(s+2)=0$$

 

[HallsofIvy]

You wrote the problem as y"- y= 0, but you solved y"- y'= 0.